TMUA Sequences and Series: Three non-zero real numbers form an arithmetic progression (in that order); their squares,…

Question

Three non-zero real numbers form an arithmetic progression (in that order); their squares, taken in the same order, form a geometric progression. What are all possible values of the common ratio of that geometric progression?

Lemma · Accepted Answer

Write the three terms of the AP in symmetric form: $\;a - d,\, a,\, a + d\;$ with $a \ne 0$ . (Symmetric choice simplifies the algebra.)

Their squares are $\;(a - d)^{2},\, a^{2},\, (a + d)^{2}$ .

For these to be three consecutive terms of a GP, the middle term squared equals the product of the outers:

\left(a^{2}\right)^{2} \;=\; (a - d)^{2}\,(a + d)^{2} \;=\; \bigl((a - d)(a + d)\bigr)^{2} \;=\; \bigl(a^{2} - d^{2}\bigr)^{2}.

So $\;a^{4} = (a^{2} - d^{2})^{2}$ , which gives $\;a^{2} = \pm (a^{2} - d^{2})$ .

Case 1: $a^{2} = a^{2} - d^{2}$ gives $d^{2} = 0$ , hence $d = 0$ . The AP is constant, all three numbers equal $a$ . Their squares are all $a^{2}$ , a GP of common ratio $1$ . But Dorofeev’s problem requires three distinct terms (otherwise the AP/GP framing is degenerate); also, the common ratio $1$ is the trivial case that occurs for any constant sequence and is usually excluded.
Case 2: $a^{2} = -(a^{2} - d^{2})$ gives $2 a^{2} = d^{2}$ , hence $d = \pm a\sqrt{2}$ .

Compute the common ratio $r = a^{2}/(a - d)^{2}$ in each sub-case.

Sub-case $d = a\sqrt{2}$ : $\;a - d = a(1 - \sqrt{2})$ , so $(a - d)^{2} = a^{2}(1 - \sqrt{2})^{2} = a^{2}(3 - 2\sqrt{2})$ . Therefore

r \;=\; \frac{a^{2}}{a^{2}(3 - 2\sqrt{2})} \;=\; \frac{1}{3 - 2\sqrt{2}} \;=\; \frac{3 + 2\sqrt{2}}{(3)^{2} - (2\sqrt{2})^{2}} \;=\; \frac{3 + 2\sqrt{2}}{9 - 8} \;=\; 3 + 2\sqrt{2}.

Sub-case $d = -a\sqrt{2}$ : by the same calculation with $1 - \sqrt{2}$ replaced by $1 + \sqrt{2}$ ,

r \;=\; \frac{1}{(1 + \sqrt{2})^{2}} \;=\; \frac{1}{3 + 2\sqrt{2}} \;=\; 3 - 2\sqrt{2}.

Both values are positive (since $3 > 2\sqrt{2} \approx 2.83$ ), so they are valid GP common ratios. The two possible common ratios are $\;r = 3 + 2\sqrt{2}\;$ and $\;r = 3 - 2\sqrt{2}$ . This is option D.

(Sanity check: $\;(3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 9 - 8 = 1$ , so the two ratios are reciprocals — consistent with the symmetry $d \to -d$ that reverses the order of the AP, which reverses the GP and inverts its common ratio.)

Why the other options fail.

A, $1$ : only the degenerate constant case, usually excluded.
B, $2$ and $\tfrac{1}{2}$ : the right idea (the ratios are reciprocals) but the wrong numbers.
C, $\sqrt{2}$ : arises from $d = a\sqrt{2}$ if a student computes $|d|/|a|$ instead of the GP ratio.
E: a valid non-degenerate progression exists, e.g. $1,\, 1 + \sqrt{2},\, 1 + 2\sqrt{2}$ in AP whose squares are $1,\, 3 + 2\sqrt{2},\, 9 + 4\sqrt{2} + 8 = 17 + 12\sqrt{2}$ , and indeed $(3 + 2\sqrt{2})^{2} = 9 + 12\sqrt{2} + 8 = 17 + 12\sqrt{2}$ ✓.

The lesson: the symmetric form $a - d,\, a,\, a + d$ for three terms of an AP makes the algebra cleaner. The condition ‘middle squared = product of outers’ characterises GP triples. The two valid common ratios are reciprocals.